Chapter 6. Mutually Orthogonal Latin Squares
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120
6. Mutually Orthogonal Latin Squares
b) L 3 , L 4 and L 5 are three mutually orthogonal latin squares of order 4, where
L3 =
1
3
4
2
4
2
1
3
2
4
3
3
1
2
and L 5 =
1
4
2
3
3
2
4
1
1
4
1
3
2
4
2
3
1
4
L4 =
1
2
3
4
2
1
4
3
3
4
1
2
4
3
2
1
.
The study of pairs of orthogonal latin squares goes back to 1782 when Euler [9]
considered the following problem.
The Euler Officer Problem. Six officers from each of six different regiments are selected so that the six officers from each regiment are of six
different ranks, the same six ranks being represented by each regiment.
Is it possible to arrange these 36 officers in a 6×6 array so that each regiment and each rank is represented exactly once in each row and column
of this array?
If we number the ranks 1, 2, 3, 4, 5 and 6, and number the regiments 1, 2,
3, 4, 5 and 6, then each officer is represented by a unique ordered pair (x, y) ∈
{1, 2, 3, 4, 5, 6} × {1, 2, 3, 4, 5, 6}, the first coordinate being his rank, the second his
regiment. Using this representation it is clear that Euler’s Officer Problem asks if it
is possible to arrange the 36 ordered pairs (x, y); x, y ∈ {1, 2, 3, 4, 5, 6}; on a 6 × 6
grid so that in each row and column the first coordinates are 1, 2, 3, 4, 5 and 6 in
some order and the second coordinates are 1, 2, 3, 4, 5, 6 in some order.
Clearly, if a solution to the Euler Officer Problem exists, the first coordinates form
a latin square and the second coordinates form a latin square (Figure 6.3). Hence the
Euler Officer Problem is equivalent to a pair of orthogonal latin squares of order 6.
However in 1900 G. Tarry [29] used brute force to show that no such pair of
orthogonal latin squares exists! For a non-brute force proof see Doug Stinson [28].
A latin square of order n is in standard form if for 1 ≤ i ≤ n cell (1, i ) contains
the symbol i . If L 1 and L 2 are orthogonal, and if L 1 and L 2 are produced from
L 1 and L 2 respectively by renaming the symbols, then clearly L 1 and L 2 are also
orthogonal. (See Exercise 6.1.2.) Therefore, if we have a set of mutually orthogonal
latin squares of order n (MOLS(n)), we can always rename the symbols in each latin
square so that each of the squares in the resulting set of MOLS(n) is in standard
form.
6.1. Introduction
121
11 12
13 14 15 16
21 22
23 24 25 26
31 32
33 34 35 36
41 42
43 44 45 46
51 52
53 54 55 56
61 62
63 64 65 66
−→
Figure 6.2: The Euler Officer Problem: can the 36 ordered pairs formed from
1, 2, 3, 4, 5, 6 be arranged on a 6 × 6 grid so that the first coordinates are different in each row and column and the second coordinates are different in each row and
each column?
If
is a solution of the
Euler Officer Problem,
then
First
Coordinates
and
Second
Coordinates
form a pair of orthogonal latin squares.
Figure 6.3: Solution of the Euler Officer Problem.
122
6. Mutually Orthogonal Latin Squares
If {L 1 , . . . , L t } is a set of MOLS(n), it is natural to ask how large t can be. Since
we can assume that each of L 1 , . . . , L t is in standard form, we can quickly obtain an
upper bound on t by considering the symbols in cell (2, 1) of L 1 , . . . , L t (so we are
assuming that n ≥ 2).
Now, the symbol 1 occurs in cell (1, 1) of each of these latin squares, so the cell
(2, 1) in each of them must be occupied by a symbol from the set
{2, 3, 4, . . . , n}. Furthermore, when any two of these latin squares are superimposed,
since they are in standard form, the first rows give the ordered pairs (1, 1), (2, 2),
(3, 3), (4, 4), . . . , (n, n). Hence the cell (2, 1) cannot be occupied by the same symbol x in different squares since the ordered pair (x, x) would occur twice when they
are superimposed (see Figure 6.4). So the symbols in cell (2, 1) of L 1 , . . . , L t are
all different and belong to the set {2, 3, . . . , n}. Therefore, if n ≥ 2 then t ≤ n − 1.
Clearly this upper bound can be obtained, since 2 MOLS(3) and 3 MOLS(4) are
exhibited in Example 6.1.1. A complete set of MOLS(n) is defined to be a set of
n − 1 MOLS(n). In the next section we will give a construction for complete sets of
MOLS(n) when n is a prime power.
1
2
3
n
L1
1
2
3
n
L2
1
2
3
n
Lt
Figure 6.4: MOLS(n) in standard form. The cell (2, 1) must be occupied by a symbol
from the set {2, 3, . . . , n} and no two squares can have the same symbol in cell (2,
1).
Exercises
6.1.2 If L is a latin square of order n and α is any permutation on {1, 2, 3, . . . , n},
denote by Lα the latin square obtained from L by replacing each symbol x in L by
xα. Prove that if L 1 and L 2 are orthogonal and α and β are any two permutations,
then L 1 α and L 2 β are also orthogonal.
6.1.3 Apply a permutation to the symbols in L 3 , L 4 and L 5 of Example 6.1.1 to
produce a complete set of MOLS(4) in standard form.
6.1.4 Show that if {L 1 , . . . , L t } is a set of MOLS(n), all of which are idempotent,
then t ≤ n − 2. (Notice that 2 of the MOLS(4) in Example 6.1.1 are idempotent, so
this bound can be attained.)
6.2. The Euler and MacNeish Conjectures
123
6.1.5 Show that any set of n − 2 MOLS(n) can be extended to a complete set of
MOLS(n).
6.2 The Euler and MacNeish Conjectures
As mentioned in Section 5.1, in 1782 Euler considered the problem of finding two
orthogonal latin squares of order 6. At that time he made the following conjecture.
The Euler Conjecture A pair of orthogonal latin squares of order n exists if and
only if n ≡ 0, 1, or 3 (mod 4).
Euler made this conjecture because he was able to construct a pair of orthogonal
latin squares of every order n ≡ 0, 1, or 3 (mod 4) and was not able to construct such
a pair for any n ≡ 2 (mod 4).
In 1900 G. Tarry [29] proved that there did not exist a pair of orthogonal latin
squares of order 6. (Tarry used brute force. For a nonbrute-force proof see Doug
Stinson [28].)
In 1922 H. MacNeish [20] made the following conjecture.
r
r
The MacNeish Conjecture Let n = p11 p22 · · · prxx , where each of p1 , p2 , p3 , . . . ,
p x is a distinct prime. Then the maximum number of MOLS(n) is m(n) =
r
min{ p1r1 , p2r2 , p33 , . . . , prxx } − 1.
Clearly the Euler Conjecture is a special case of the MacNeish Conjecture, since
r r
if n ≡ 2 (mod 4), n = 2· (odd number) = 2 · p22 p33 · · · prxx , and so m(n) =
r3
r2
rx
min{2, p2 , p3 , . . . , px } − 1 = 2 − 1 = 1.
Euler Conjecture
MacNeish Conjecture
Figure 6.5: The Euler Conjecture is a subcase of the MacNeish Conjecture.
MacNeish had a very good reason for his conjecture. He found a finite field construction and a direct product construction which enabled him to construct m(n)
MOLS(n) for every n. (Finite fields had not been invented in Euler’s day.)
124
6. Mutually Orthogonal Latin Squares
The first inkling that the Euler Conjecture might not be true came in 1957 when E.
T. Parker constructed 3 MOLS(21), thereby killing the MacNeish Conjecture. Since
the Euler Conjecture is a special case of the MacNeish Conjecture it was not particularly surprising when, in 1958, R. C. Bose and S. S. Shrikhande [2] constructed a pair
of orthogonal latin squares of order 22, and in so doing disproved the long standing
Euler Conjecture. Shortly after this “event,” E. T. Parker [22] constructed a pair of
orthogonal latin squares of order 10. Finally, in 1960 R. C. Bose, S. S. Shrikhande,
and E. T. Parker [3] proved that a pair of orthogonal latin squares of order n exists
for all n other than 2 or 6 (for which no such pair exists).
In what follows, we will give a systematic verification of all of the above existence
results. It will be convenient to use quasigroup notation to define orthogonal latin
squares in much of what follows.
Everything that follows is predicated on a reasonable knowledge of finite fields.
Among other things, how to construct them, as well as the fact that the spectrum (=
the set of all orders for which a finite field exists) is precisely the set of all pr , where
p is a prime and r is a positive integer.
The Finite Field Construction Let (F, +, ·) be a finite field of order n. Since
there is nothing sacred about the symbols representing the field elements in F, it is
convenient to assume that F = {1, 2, 3, . . . , n} with the proviso the n represents the
zero field element. For each n = k ∈ F, define a binary operation ◦(k) on F by
x ◦ (k)y = x · k + y (arithmetic computed in the finite field (F, +, ·). Then (F, ◦(k))
is a quasigroup (latin square).
It is easy to see that (F, ◦(k)) is in fact a quasigroup. Suppose x ◦ (k)y = x ◦ (k)z.
Then x · k + y = x · k + z and so y = z. On the other hand if y ◦ (k)x = z ◦ (k)x,
then y · k + x = z · k + x, so y · k = z · k, and since k = zero we get y = z.
Example 6.2.1 A quasigroup of order 4 constructed using the Finite Field Construction. We first construct the finite field of order 4 using the irreducible polynomial
1 + x + x 2.
+
0
1
x
1+x
·
0
1
x
1+x
0
0
1
x
1+x
0
0
0
0
0
1
1
0
1+x
x
1
0
1
x
1+x
x
x
1+x
0
1
x
0
x
1+x
1
x
1
0
1+x
0
1+x
1
x
1+x 1+x
Renaming the field elements 0, 1, x, and 1 + x with 4, 1, 2, and 3 respectively
gives the tables
6.2. The Euler and MacNeish Conjectures
125
+
4
1
2
3
·
4
1
2
3
4
4
1
2
3
4
4
4
4
4
1
1
4
3
2
1
4
1
2
3
2
2
3
4
1
2
4
2
3
1
3
3
2
1
4
3
4
3
1
2
Since there are 3 nonzero field elements 1, 2, and 3, we can construct 3 quasigroups as follows:
◦(1)
4
1
2
3
◦(2)
4
1
2
3
4
4
1
2
3
4
4
1
2
3
1
1
4
3
2
1
2
3
4
1
2
2
3
4
1
2
3
2
1
4
3
3
2
1
4
3
1
4
3
2
x ·1+y
x ·2+y
◦(3)
4
1
2
3
4
4
1
2
3
1
3
2
1
4
2
1
4
3
2
3
2
3
4
1
x ·3+y
(Using the addition and multiplication tables of the finite field of order 4 given
above.)
Theorem 6.2.2 Let (F, +, ·) be a finite field of order n, where F = {1, 2, 3,
. . . , n = zero}. Then the quasigroups (F, ◦(1)), (F, ◦(2)), . . . , (F, ◦(n − 1)) are a
complete set of mutually orthogonal quasigroups (latin squares) of order n.
126
6. Mutually Orthogonal Latin Squares
Proof. It is convenient to use latin square vernacular in the proof. We denote the
latin square associated with the quasigroup (F, ◦(k)) by L(k). We will check that
L(k) and L( ) are orthogonal for 1 ≤ k < < n by using The Two-Finger Rule.
Suppose that cells (i 1 , j1) and (i 2 , j2 ) both contain the same symbol in L(k) and
both contain the same symbol in L( ). We will prove that i 1 = i 2 and j1 = j2 . For
convenience, let x · y be denoted by x y.
By definition of (F, ◦(k)) and (F, ◦( )) we have that
i 1 k + j1 = i 2 k + j2 , and
i 1 + j1 = i 2 + j2 .
So
i 1 k + j1 − i 2 k = j2 = i 1 + j1 − i 2 , or
(i 1 − i 2 )k = (i 1 − i 2 ) .
Since k = and k, = zero, this means that i 1 − i 2 = zero, so i 1 = i 2 and therefore
j1 = j2 .
Example 6.2.3 The 3 quasigroups in Example 6.2.1 constructed using The Finite
Field Construction are mutually orthogonal (check it out!).
Example 6.2.4 We construct 4 MOLS(5) using the finite field construction. The
finite field of order 5 is, of course, the integers (mod 5). So we can easily get the
following 4 orthogonal quasigroups of order 5 by defining i ◦ (k) j = i k + j (mod
5), renaming zero with 5.
◦(1) 5
1
2
3
4
◦(2) 5
1
2
3
4
5
5
1
2
3
4
5
5
1
2
3
4
1
1
2
3
4
5
1
2
3
4
5
1
2
2
3
4
5
1
2
4
5
1
2
3
3
3
4
5
1
2
3
1
2
3
4
5
4
4
5
1
2
3
4
3
4
5
1
2
i ·1+ j
i ·2+ j
6.2. The Euler and MacNeish Conjectures
◦(3) 5
1
2
3
4
127
◦(4) 5
1
2
3
4
5
5
1
2
3
4
5
5
1
2
3
4
1
3
4
5
1
2
1
4
5
1
2
3
2
1
2
3
4
5
2
3
4
5
1
2
3
4
5
1
2
3
3
2
3
4
5
1
4
2
3
4
5
1
4
1
2
3
4
5
i ·3+ j
i ·4+ j
Example 6.2.5 The finite field (F, +, ·) of order 25 can be constructed using the
irreducible primitive polynomial p(x) = 3 + 2x + x 2 . To facilitate the arithmetic
in (F, +, ·), we represent each non-zero field element as a power of the primitive
element x as follows:
i
0
1
2
3
4
5
xi
1
2
1
2
3
+
+
+
+
x
3x
x
4x
4x
i
6
7
8
9
10
11
xi
3
1
3
1
4
+
+
+
+
3x
4x
3x
2x
2x
i
12
13
14
15
16
17
xi
4
3
4
3
2
+
+
+
+
4x
2x
4x
x
x
i
18
19
20
21
22
23
xi
2
4
2
4
1
+
+
+
+
2x
x
2x
3x
3x
a) Find the cell in “column” x 9 of (F, ◦(x 19 )) that contains the symbol x 11 .
By definition, cell (i, j ) of (F, ◦(k)) contains symbol i k + j . We need to find
the row i , and we know that
i x 19 + x 9 = x 11 , so
i x 19 = x 11 − x 9
= (4 + 2x) − (3 + 3x)
= 1 + 4x, so
i = x 8 /x 19
= x 13
(since x 24 = 1, so x −11 = x −11 x 24 = x 13 ). Therefore cell (x 13 , x 9 ) of
(F, ◦(x 19 )) contains symbol x 11 .
b) Find the cell that contains the symbol x 20 in (F, ◦(x 6 )) and the symbol x 10 in
(F, ◦(x 14 )).
128
6. Mutually Orthogonal Latin Squares
Now we have to find i and j , where
i x 6 + j = x 20 , and
i x 14 + j = x 10 .
Solving these simultaneously, we get
i (x 14 − x 6 ) = x 10 − x 20 , so
i = ((1 + 2x) − (4 + x))/((3 + 2x) − (3))
= (2 + x)/2x
= x 17 /x 19
= x 22 ,
(again, since x 24 = 1). Substituting for i gives
x 22 x 6 + j = x 20 , so
j = x 20 − x 28
= (4 + x) − (2 + 4x)
= x 21 .
Therefore we have that cell (x 22 , x 21 ) contains the symbol x 20 in (F, ◦(x 6 )) and
contains the symbol x 10 in (F, ◦(x 14 )).
Exercises
6.2.6 Form addition and multiplication tables for the finite field of order 7. Use these
to construct 6 MOLS(7).
6.2.7 Form addition and multiplication tables for the finite field of order 8, constructing the field with the irreducible polynomial p(x) = 1 + x + x 3 . Rename the
field elements with the symbols 1, . . . , 8 and construct 7 MOLS(8).
6.2.8 Form addition and multiplication tables for the finite field of order 9, constructing the field with the irreducible polynomial p(x) = 2 + 2x + x 2 . Rename the
field elements with the symbols 1, . . . , 9 and construct 8 MOLS(9).
6.2.9 Continuing Example 6.2.5(a), find the cell in
(a) row x 3 of L(x 12 ) that contains symbol x 15 ,
(b) row x 4 of L(x 16 ) that contains symbol x 0 = 1,
(c) column 0 of L(x 9 ) that contains symbol x 7 , and
(d) column x 19 of L(x 21 ) that contains symbol x 23 .
6.2. The Euler and MacNeish Conjectures
129
6.2.10 Continuing Example 6.2.5(b), find the cell that contains
(a) symbol x 4 in L(x 4 ) and symbol x 8 in L(x 10 ),
(b) symbol x 0 in L(x 2 ) and symbol x 14 in L(x 20 ), and
(c) symbol x 3 in L(x 3 ) and symbol 0 in L(x 18 ).
6.2.11 Let L be a latin square of order n and let r be any permutation on
{1, 2, 3, . . . , n}. Denote by r L the latin square obtained from L by permuting the
rows of L according to r . Let L 1 , L 2 , . . . , L t be t MOLS(n) and r any permutation
of the rows of L 1 so that r L 1 has a constant main diagonal. Then the latin squares
r L 1 , r L 2 , . . . , r L t are t MOLS(n) and furthermore each of r L 2 , r L 3 , . . . , r L t is
diagonalized (= no 2 symbols on the main diagonal are the same). Applying appropriate permutations α2 , α3 , . . . , αt to the symbols in each of r L 2 , r L 3 , . . . , r L t
gives t − 1 idempotent MOLS(n): r L 2 α2 , r L 3 α3 , . . . , r L t αt . Use the above algorithm to construct 2 idempotent MOLS(4), 3 idempotent MOLS(5), and 5 idempotent MOLS(7). (See Examples 6.2.1 and 6.2.4 and Exercise 6.2.6 respectively.)
6.2.12 Prove that there exists m(n) − 1 idempotent MOLS(n).
We now present the direct product construction.
The Direct Product Construction The direct product A × B of two latin squares
A and B of orders m and n respectively is the latin square of order mn defined by
A×B
A(1,1) A(1,2)
A(1,n)
A(2,1) A(2,2)
A(2,n)
A(n,1) A(n,2)
A(n,n)
=
where A(i, j ) is the latin square of order m formed from A by replacing each entry
x with (x, k), where k is the entry in cell (i, j ) of B.
Example 6.2.13 The direct product of latin squares of orders 3 and 4.
130
6. Mutually Orthogonal Latin Squares
A=
3
1
2
2
3
1
1
2
3
1
3
4
2
4
2
1
3
2
4
3
1
3
1
2
4
B=
31 11 21 33 13 23 34 14 24 32 12 22
21 31 11 23 33 13 24 34 14 22 32 12
11 21 31 13 23 33 14 24 34 12 22 32
34 14 24 32 12 22 31 11 21 33 13 23
24 34 14 22 32 12 21 31 11 23 33 13
14 24 34 12 22 32 11 21 31 13 23 33
A×B=
32 12 22 34 14 24 33 13 23 31 11 21
22 32 12 24 34 14 23 33 13 21 31 11
12 22 32 14 24 34 13 23 33 11 21 31
33 13 23 31 11 21 32 12 22 34 14 24
23 33 13 21 31 11 22 32 12 24 34 14
13 23 33 11 21 31 12 22 32 14 24 34
The direct product of the quasigroups (A, ◦1 ) and (B, ◦2 ) is just the direct product
of the latin squares A and B with appropriate headlines and sidelines. Algebraically,
the direct product is defined to be the quasigroup (A× B, ) where (x, y) (z, w) =
(x ◦1 z, y ◦2 w).
Example 6.2.14 The direct product of the quasigroups corresponding to the latin
squares A and B in Example 6.2.13.